分析
题目要求生成树个数，求出基尔霍夫矩阵后高斯消元，
但是这里模数不是质数，所以要辗转相除法
代码
#include <cstdio>
#include <cctype>
#include <cstring>
#include <algorithm>
#include <cmath>
#define rr register
using namespace std;
const int mod=1000000000; typedef long long lll;
int a[82][82],CNT,n,m,rk[82][82];
inline signed iut(){
rr int ans=0; rr char c=getchar();
while (!isdigit(c)) c=getchar();
while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();
return ans;
}
inline void Mo(int &x,int y){x=x<y?x-y+mod:x-y;}
inline signed mo(int x,int y){return x+y>=mod?x+y-mod:x+y;}
inline void add(int x,int y){++a[x][x],++a[y][y],Mo(a[x][y],1),Mo(a[y][x],1);}
inline void doit(int t1,int t2,lll &ai,lll &aj,lll &Ai,lll &Aj,int &F){
ai=Aj=1,aj=Ai=0;
while (t2){
ai-=t1/t2*Ai,aj-=t1/t2*Aj,t1%=t2,
ai=(ai%mod+mod)%mod,aj=(aj%mod+mod)%mod;
swap(t1,t2),swap(ai,Ai),swap(aj,Aj),F=mod-F;
}
}
inline signed Gauss(int n){
rr int ans=1; rr lll ai,aj,Ai,Aj;
for (rr int i=1;i<=n;++i){
for (rr int j=i+1;j<=n;++j){
rr int t1=a[i][i],t2=a[j][i];
doit(t1,t2,ai,aj,Ai,Aj,ans);
for (rr int k=1;k<=n;++k){
rr int T1=mo(a[i][k]*ai%mod,a[j][k]*aj%mod);
rr int T2=mo(a[i][k]*Ai%mod,a[j][k]*Aj%mod);
a[i][k]=T1,a[j][k]=T2;
}
}
ans=1ll*ans*a[i][i]%mod;
}
return ans;
}
signed main(){
n=iut(); m=iut();
for (rr int i=1;i<=n;++i)
for (rr int j=1;j<=m;++j){
rr char c=getchar();
while (c!='*'&&c!='.') c=getchar();
if (c=='.') rk[i][j]=++CNT;
}
for (rr int i=1;i<=n;++i)
for (rr int j=1;j<=m;++j)
if (rk[i][j]){
if (j<m&&rk[i][j+1]) add(rk[i][j],rk[i][j+1]);
if (i<n&&rk[i+1][j]) add(rk[i][j],rk[i+1][j]);
}
return !printf("%d",Gauss(CNT-1));
}